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lana [24]
3 years ago
12

If a snack cake contains 450 food calories and Carla

Chemistry
1 answer:
Oksana_A [137]3 years ago
8 0

Answer:

9/10hr

Explanation:

From the question given:

Carla was able to burn 250 calories by running for one-half (1/2)hr.

Therefore, Carla will burn 450calories by running for = (450x1/2) /250 = 225/250 = 9/10hr

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Ymorist [56]

Answer:

ok

Explanation:

6 0
3 years ago
The orbital radius of Venus is 0.72 AU. What is this distance in kilometers? (One AU is about 150 million kilometers.)
Evgen [1.6K]
<span>The answer is D) 108 million kilometers. To solve this problem, you must perform a simple unit conversion calculation. 1 AU = 150,000,000 km is the conversion factor. Take the radius of Venus, .72 AU, and multiply it by 150,000,000 km/1 AU. You flip the conversion factor so that the units of the original distance in the numerator cancel the units in the denominator of the conversion factor. completing the calculation gives you 108,000,000 km</span>
6 0
2 years ago
The pH of a solution in which the concentration of H+ is 0.010M will be:
shusha [124]

B: 2

If it's right plz give brainliest lol <3

7 0
2 years ago
You are starting with 2 pounds of 30% acetic acid solution. You want to dilute
Tpy6a [65]

Answer:

The mass of water to be added is 2 pounds

Explanation:

The given parameters are;

The mass of the given solution = 2 pounds

The concentration of the given solution = 30%

The desired concentration of the solution = 15%

The mass, m of the acetic acid in the given solution = 30% × 2 pounds

m = 30/100 × 2 pounds = 0.6 pounds

To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;

15% of X = 0.6 pounds

15/100 × X = 3/20 × 0.6 pounds

∴ The mass of the solution required  X = 0.6 × 20/3 = 4 pounds

The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds

Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.

8 0
3 years ago
Fe2O3 + 2Al -&gt; Al2O3 + 2Fe
pashok25 [27]
Fe2O3 + 2Al ---> Al2O3 + 2Fe 
Mole ratio Fe2O3 : Al = 1:2 
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles 
No. of moles of Al = 150/27 = 5.555555555 moles. 
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react. 

So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared. 
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe 
7 0
3 years ago
Read 2 more answers
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